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1.9: The Heisenberg Uncertainty Principle

Skills to Develop

  • To understand that sometime you cannot know everything about a Quantum system as demonstrated by the Heisenberg uncertainly principle.

In classical physics, studying the behavior of a physical system is often a simple task due to the fact that several physical qualities can be measured simultaneously. However, this possibility is absent in the quantum world. In 1927 the German physicist Werner Heisenberg described such limitations as the Heisenberg Uncertainty Principle, or simply the Uncertainty Principle, stating that it is not possible to measure both the momentum and position of a particle simultaneously.

The Heisenberg Uncertainty Principle is a fundamental theory in quantum mechanics that defines why a scientist cannot measure multiple quantum variables simultaneously. Until the dawn of quantum mechanics, it was held as a fact that all variables of an object could be known to exact precision simultaneously for a given moment. Newtonian physics placed no limits on how better procedures and techniques could reduce measurement uncertainty so that it was conceivable that with proper care and accuracy all information could be defined. Heisenberg made the bold proposition that there is a lower limit to this precision making our knowledge of a particle inherently uncertain.

Heisenberg’s Uncertainty Principle

It is mathematically possible to express the uncertainty that, Heisenberg concluded, always exists if one attempts to measure the momentum and position of particles. First, we must define the variable “x” as the position of the particle, and define “p” as the momentum of the particle. The momentum of a photon of light is known to simply be its frequency, expressed by the ratio \(h/λ\), where h represents Planck’s constant and \(\lambda\) represents the wavelength of the photon. The position of a photon of light is simply its wavelength (\(\lambda\)). To represent finite change in quantities, the Greek uppercase letter delta, or Δ, is placed in front of the quantity. Therefore,

\[\Delta{p}=\dfrac{h}{\lambda} \label{1.9.1}\]

\[\Delta{x}= \lambda \label{1.9.2}\]

By substituting \(\Delta{x}\) for \(\lambda\) into Equation \(\ref{1.9.1}\), we derive

\[\Delta{p}=\dfrac{h}{\Delta{x}} \label{1.9.3}\]

or,

\[\underset{\text{early form of uncertainty principle }}{\Delta{p}\Delta{x}=h} \label{1.9.4}\]

A Common Trend in Quantum Systems

Equation \(\ref{1.9.4}\) can be derived by assuming the particle of interest is behaving as a particle, and not as a wave. Simply let \(\Delta p=mv\), and \(Δx=h/(m v)\) (from De Broglie’s expression for the wavelength of a particle). Substituting in \(Δp\) for \(mv\) in the second equation leads Equation \(\ref{1.9.4}\)

This equation was further refined by Heisenberg and his colleague Niels Bohr, and was eventually rewritten as 

\[\Delta{p_x}\Delta{x} \ge \dfrac{h}{4\pi} = \dfrac{\hbar}{2} \label{1.9.5}\]

with \(\hbar = \frac{h}{2\pi}= 1.0545718 \times 10^{-34}\; m^2 \cdot kg / s\).

Equation \(\ref{1.9.5}\) reveals that the more accurately a particle’s position is known (the smaller \(Δx\) is), the less accurately the momentum of the particle in the x direction (\(Δp_x\)) is known. Mathematically, this occurs because the smaller \(Δx\) becomes, the larger \(Δp_x\) must become in order to satisfy the inequality. However, the more accurately momentum is known the less accurately position is known (Figure \(\PageIndex{1}\)). 

 

reality_wavepacket.gif

Figure \(\PageIndex{1}\): The animation shows the relevant spreads in the uncertainty for position and momentum of light/photons (light wave's corresponding photon particle). From the result of de Broglie, we know that for a particle with known momentum, \(p\) will have a precise value for its de Broglie wavelength can be determined (and hence a specific color of the light).

What is the Proper Definition of Uncertainty?

Equation \(\ref{1.9.5}\) relates the uncertainty of momentum and position. An immediate questions that arise is if \(\Delta x\) represents the full range of possible \(x\) values or if it is half (e.g., \(\langle x \rangle \pm \Delta x\)). \(\Delta x\) is the standard deviation and is a statistic measure of the spread of \(x\) values. The use of half the possible range is more accurate estimate of \(\Delta x\). Later, we can show that once a wavefunction can be constructed to describe the system, then both \(x\) and \(\Delta x\) can be explicitly derived. However, that is introduced later and the simple approximation above works fine for now.

For example: If a problem argues a particle is trapped in a box of length, \(L\), then the uncertainly of it position is \(\pm L/2\). So the value of \(\Delta x\) used in Equation \(\ref{1.9.5}\) is \(L/2\), not \(L\).

Example \(\PageIndex{1}\)

An electron is confined to the size of a magnesium atom with a 150 pm radius. What is the minimum uncertainty in its velocity?

Solution

The uncertainty principle (Equation \(\ref{1.9.5}\)):

\[\Delta{p}\Delta{x} \ge \dfrac{\hbar}{2} \nonumber\]

can be written  

\[\Delta{p} \ge \dfrac{\hbar}{2 \Delta{x}} \nonumber \]

and substituting \(\Delta p=m \Delta v \) since the mass is not uncertain.

\[\Delta{v} \ge \dfrac{\hbar}{2\; m\; \Delta{x}} \nonumber \]

the relevant parameters are

  • mass of electron \(m=m_e= 9.109383 \times 10^{-31}\; kg\)
  • uncertainty in position: \(\Delta x=150 \times 10^{-12} m \)

\[\Delta{v} \ge \dfrac{1.0545718 \times 10^{-34}  \cancel{kg} m^{\cancel{2}} / s}{(2)\;( 9.109383 \times 10^{-31} \; \cancel{kg}) \; (150 \times 10^{-12} \; \cancel{m}) } = 3.9 \times 10^5\; m/s \nonumber\]

Exercise \(\PageIndex{1}\)

What is the maximum uncertainty of velocity the electron described in Example \(\PageIndex{1}\)?

Answer

Infinity. There is no limit in the maximum uncertainty, just the minimum uncertainty.

Understanding the Uncertainty Principle through Wave Packets and the Slit Experiment

It is hard for most people to accept the uncertainty principle, because in classical physics the velocity and position of an object can be calculated with certainty and accuracy. However, in quantum mechanics, the wave-particle duality of electrons does not allow us to accurately calculate both the momentum and position because the wave is not in one exact location but is spread out over space. A "wave packet" can be used to demonstrate how either the momentum or position of a particle can be precisely calculated, but not both of them simultaneously. An accumulation of waves of varying wavelengths can be combined to create an average wavelength through an interference pattern: this average wavelength is called the "wave packet". The more waves that are combined in the "wave packet", the more precise the position of the particle becomes and the more uncertain the momentum becomes because more wavelengths of varying momenta are added. Conversely, if we want a more precise momentum, we would add less wavelengths to the "wave packet" and then the position would become more uncertain. Therefore, there is no way to find both the position and momentum of a particle simultaneously. 

Several scientists have debated the Uncertainty Principle, including Einstein. Einstein created a slit experiment to try and disprove the Uncertainty Principle. He had light passing through a slit, which causes an uncertainty of momentum because the light behaves like a particle and a wave as it passes through the slit. Therefore, the momentum is unknown, but the initial position of the particle is known. Here is a video that demonstrates particles of light passing through a slit and as the slit becomes smaller, the final possible array of directions of the particles becomes wider. As the position of the particle becomes more precise when the slit is narrowed, the direction, or therefore the momentum, of the particle becomes less known as seen by a wider horizontal distribution of the light. 

Example \(\PageIndex{2}\)

The speed of a 1.0 g projectile is known to within \(10^{-6}\;m/s\). Calculate the minimum uncertainty in its position. What is the maximum uncertainty of its position?

Solution

From Equation \(\ref{1.9.5}\), the \(\Delta{p_x} = m \Delta v_x\) with \(m=1.0\;g\). Solving for \(\Delta{x}\) to get

\[ \Delta{x} = \dfrac{\hbar}{2m\Delta v} \nonumber\]

\[\Delta{x} = \dfrac{1.0545718 \times 10^{-34} \; m^2 \cdot kg / s}{(2)(1.0 \; g)(10^{-6} \;m/2)} = 5 \times 10^{-26} \,m \nonumber\]

This negligible for all intents and purpose as expected for a macroscopic object.

Exercise \(\PageIndex{2}\)

Estimate the minimum uncertainty in the speed of an electron confined to a hydrogen atom within a diameter of \(1 \times  10^{-10} m\)?

Answer

\[\Delta v=1.1 \times ^6 km/s\]

Notice that the uncertainty is greater for the electron in a hydrogen atom than in the magnesium atom (Example \(\PageIndex{1}\)) as expected since the magnesium atom is appreciably bigger.

Heisenberg’s Uncertainty Principle not only helped shape the new school of thought known today as quantum mechanics, but it also helped discredit older theories. Most importantly, the Heisenberg Uncertainty Principle made it obvious that there was a fundamental error in the Bohr model of the atom. Since the position and momentum of a particle cannot be known simultaneously, Bohr’s theory that the electron traveled in a circular path of a fixed radius orbiting the nucleus was obsolete. Furthermore, Heisenberg’s uncertainty principle, when combined with other revolutionary theories in quantum mechanics, helped shape wave mechanics and the current scientific understanding of the atom.

Note: Heisenberg and the Police

  • Heisenberg get pulled over for speeding by the police. The officer asks him "Do you know how fast you were going?"
  • Heisenberg replies, "No, but we know exactly where we are!"
  • The officer looks at him confused and says "you were going 108 miles per hour!"
  • Heisenberg throws his arms up and cries, "Great! Now we're lost!"

Contributors

  • Sarah Woods, Kris Baumgartner (UC Davis)